In the circuit shown in figure,

Options

A. `i = 5/3 A` when `S_1` is closed and `S_2` is open
B. `i=5A` when `S_1` is open `S_2` is closed
C. `i=2A` when `S_1` and `S_2` both are open
D. `i=10` A when both `S_1` and `S_2` are closed

Answer

B, C, D

Explanation

b., c., d.
When `S_2` is closed and `S_2` is open
`I = (20-10)/3 = 10/3A`
When `S_1` is open and `S_2` is closed
Taking outer loop
`-2I -2I + 20 = 0`
or `I = 20/4 A = 5A`
When `S_1` and `S_2` both are open
`I = (20-10)/5 = 2A`



When both are closed

Again taking outer loop `I = 20/2 = 10A`.